Answer: i)$L^{-1}\left\{{tan}^{-1}{\left(s+2\right)}^2\right\} $

$\phi(s)=tan^{-1}(s+2)^2$

Differentiating with respect to s we get , 

$\phi'(s)=\dfrac{2(s+2)}{(s+2)^2+1}$

Taking laplace inverse on $\phi(s)$

$L^{-1}[\phi(s)]=-\dfrac{1}{t}L^{-1}[\phi'(s)]$

$\therefore L^{-1}[\phi(s)]=-\dfrac{1}{t}L^{-1}[\dfrac{2(s+2)}{(s+2)^2+1}]$

$\therefore L^{-1}[\phi(s)]=-\dfrac{2}{t}L^{-1}[\dfrac{(s+2)}{(s+2)^2+1}]$

$\therefore L^{-1}[tan^{-1}(s+2)^2]=-\dfrac{2}{t}e^{-2t}cost$

ii) $L\{t^2H(t-2)-cosht \delta(t-4)\}$

a) Here $f(t)=t^2 \ and \ a=2$

$\therefore f(t+2)=(t+2)^2=t^2+4t+4$

$L[f(t+2)]=L[t^2+4t+4]=\dfrac{2}{s^3}+4\cdot \dfrac{1}{s^2} + \cdot \dfrac{1}{s}$

$L[t^2H(t-2)]=e^{-2s}[\dfrac{2}{s^3}+ \dfrac{4}{s^2} + \dfrac{4}{s}]$

b) Here , $f(t)=cosht \ and \ a=4$

$\therefore L[f(t)\delta(t-a)]=e^{-as}f(a)=e^{-4s}cosh4$

Hence , $L\{t^2H(t-2)-cosht \delta(t-4)\}=e^{-2s}[\dfrac{2}{s^3}+\dfrac{4}{s^2}+\dfrac{4}{s}]-e^{-4s}cosh4$